HClO is a weak acid ( 𝐾a=4.0×10−8 ) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.020 M in NaClO at 25 °C?

Answer 1

You will want to look at the hydrolysis of the salt

NaClO + H₂O ==> NaOH + HClO .. molecular equation

ClO^- + H₂O ==> HClO + OH^- .. net ionic equation

In this reaction, ClO^- is acting as a base, so we will want to write the Kb expression:

Kb = [HClO][OH^-] / [ClO^-]

We obtain Kb from the Ka as follows: Kb = 1x10^-14 / Ka = 1x10^-14 / 4.0x10^-8 = 2.5x10^-7

2.5x10^-7 = [HClO][OH^-] / [ClO^-] = (x)(x) / 0.02-x (assume x is small relative to 0.02 and ignore it)

2.5x10^-7 = x² / 0.02

x² = 5x10^-9

x = [OH^-] = 7.1x10^-5 (note: this is significantly less than 0.02 so above assumption was valid)

Now that we have the [OH^-], we can find pOH and then pH

pOH = -log 7.1x10^-5

pOH = 4.15

pH = 14 - 4.15

pH = 9.85